MinekPo1 [it/she]

depends on your format ? if the format is binary anyway or has binary blobs (ie it needs a program that is able to handle octets outside the printable range) and using those characters does not introduce any ambiguities with the format then go for it . ANSI and related control codes all start with 0x1B

his opsec was trash . he left tons of DNA evidence and he was wearing a very specific backpack making it easy for cops to track him , once he ditched it he had enough cameras behind him that he appeared on that he could have been tracked further plus the showing his face while flirting with a receptionist . see this video by The Hated One (youtube.com , use your preferred privacy friendly front end)

yeah no . sorry but the guy was not as well prepared as it seems , he left a lot of tracks and made a lot of mistakes .

doubt it , since they shorten their username to CY78 , for example on their youtube channel profile or in the vaguely lewd unicode art

also : https://github.com/Kingcy78/NEW/blob/main/1#L551-L570

high quality malware !

the other mentioned repo has the same payload soooo

here is the extracted payload : https://gist.github.com/MinekPo1/af9bfd787c35ea5ff8b22165e9a05a6d

I suspect that’s not the actual payload , the anggur- repo appears to be more suspicious , might try to analyse that

the frames definetly can impact how thick the lenses look . my previous glasses looked quite thick but despite my prescription being higher the new ones look not as thick

unity kinda does this ! there is a json file wirh a base-64 encoded string (UTF-16 encoding) and is a completely different data structure with embedded json

honestly while I agree that slightly longer keys wont be safe for long , but tbh I’m gonna sit a bit more on my 23-bit RSA keys before migrating

it seems like it , but I have no idea !

edit : to clarify , I believe it does , using the CC1 & CC2 pins which are also used for other things , but I don’t know anything about USB protocol side , I should learn about it haha

more expensive imo.

actually the same pins (well one of them , though since the connector is rotationally symmetric you need two anyway) are used for USB Power Delivery and to negotiate what speed regime to operate in .

furthermore , USB On the go , which was introduced in USB 2.0 , offers the same functionality for USB Micro and USB Mini

actually they would be correct :

USB began as a protocol where one side (USB-A) takes the leading role and the other (USB-B) the following role . this was mandated by hardware with differently shaped plugs and ports . this made sense for the time as USB was ment to connect computers to peripherals .

however some devices don’t fit this binary that well : one might want to connect their phone to their computer to pull data off it , but they also might want to connect a keyboard to it , with the small form factor not allowing for both a USB-A and USB-B port. the solution was USB On-The-Go : USB Mini-A/B/AB and USB Micro-A/B/AB connectors have an additional pin which allows both modes of operations

with USB-C , aside from adding more pins and making the connector rotationally symmetric , a very similar yet differently named feature was included , since USB-C - USB-C connections were planed for

so yeah USB-A to USB-A connections are explicitly not allowed , for a similar reason as you only see CEE 7 (fine , or the objectively worse NEMA) plugs on both ends of a cable only in joke made cables . USB-C has additional hardware to support both sides using USB-C which USB-A , neither in the original or 3.0 revision , has .

cocaine shark , doo-doo , doo-doo , doo-doo cocaine shark , doo-doo , doo-doo , doo-doo cocaine shark , doo-doo , doo-doo , doo-doo cocaine shark !

yeah thats the point , the cybertruck design is retrofuturism

Agh I made a mistake in my code:

if (recalc || numbers[i] != (hashstate[i] & 0xffffffff)) {
	hashstate[i] = hasher.hash(((uint64_t)p << 32) | numbers[i]);
}

Since I decided to pack the hashes and previous number values into a single array and then forgot to actually properly format the values, the hash counts generated by my code were nonsense. Not sure why I did that honestly.

Also, my data analysis was trash, since even with the correct data, which as you noted is in a lineal correlation with n!, my reasoning suggests that its growing faster than it is.

Here is a plot of the incorrect ratios compared to the correct ones, which is the proper analysis and also clearly shows something is wrong.

Anyway, and this is totally unrelated to me losing an internet argument and not coping well with that, I optimized my solution a lot and turns out its actually faster to only preform the check you are doing once or twice and narrow it down from there. The checks I’m doing are for the last two elements and the midpoint (though I tried moving that about with seemingly no effect ???) with the end check going to a branch without a loop. I’m not exactly sure why, despite the hour or two I spent profiling, though my guess is that it has something to do with caching?

Also FYI I compared performance with -O3 and after modifying your implementation to use sdbm and to actually use the previous hash instead of the previous value (plus misc changes, see patch).

you forgot about updating the hashes of items after items which were modified , so while it could be slightly faster than O((n!×n)²) , not by much as my data shows .

in other words , every time you update the first hash you also need to update all the hashes after it , etcetera

so the complexity is O(n×n + n×(n-1)×(n-1)+…+n!×1) , though I dont know how to simplify that

actually all of my effort was wasted since calculating the hamming distance between two lists of n hashes has a complexity of O(n) not O(1) agh

I realized this right after walking away from my devices from this to eat something :(

edit : you can calculate the hamming distance one element at a time just after rehashing that element so nevermind

honestly I was very suspicious that you could get away with only calling the hash function once per permutation , but I couldn’t think how to prove one way or another.

so I implemented it, first in python for prototyping then in c++ for longer runs… well only half of it, ie iterating over permutations and computing the hash, but not doing anything with it. unfortunately my implementation is O(n²) anyway, unsure if there is a way to optimize it, whatever. code

as of writing I have results for lists of n ∈ 1 … 13 (13 took 18 minutes, 12 took about 1 minute, cant be bothered to run it for longer) and the number of hashes does not follow n! as your reasoning suggests, but closer to n! ⋅ n.

link for the desmos page

anyway with your proposed function it doesn’t seem to be possible to achieve O(n!²) complexity

also dont be so negative about your own creation. you could write an entire paper about this problem imho and have a problem with your name on it. though I would rather not have to title a paper “complexity of the magic lobster party problem” so yeah